For squeeze theorem, if $0 \leqslant f(x) \leqslant 1$, can this be used to prove the assertion that the limit $f(x)$ does not exist? I.e. $f(x)$ is $xy/ (x^2+xy+y^2)$, as $(x,y)\rightarrow(0,0)$.
I'm aware you can use the method of approaching from different paths, but was wondering if squeeze theorem was enough? My first guess is obviously this isn't enough to prove anything, since $f(x)$ can still be $0$, but of all the examples I tried, when the squeeze theorem doesn't squeeze $f(x)$ into a number $L$, the limit doesn't exist. Just a coincidence?
Best Answer
As I know the squeeze theorem, it says if (blah) then the limit is $L$. It says nothing if you can't prove (blah). Maybe there is a better thing to squeeze it, maybe there is no limit.